Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,\,0),$ the latus rectum is of length $12$
Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,\,0),$ the latus rectum is of length $12$
Foci $(\pm 4,\,0),$ the latus rectum is of length $12$
Here, the foci are on the $x-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
since the foci are $(\pm 4,\,0)$, $c=4$
Length of latus rectum $=12$
$\Rightarrow \frac{2 b^{2}}{a}=12$
$\Rightarrow b^{2}=6 a$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore a^{2}+6 a=16$
$\Rightarrow a^{2}+6 a-16=0$
$\Rightarrow a^{2}+8 a-2 a-16=0$
$\Rightarrow(a+8)(a-2)=0$
$\Rightarrow a=-8,2$
since a is non-negative, $a=2$
$\therefore b^{2}=6 a=6 \times 2=12$
Thus, the equation of the hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$
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